lnx = x^3 + 1
i know x=1 but can you solve this algebraically?
ln(1)=0, which does not equal 0+1
ln(x)-x^3=1 does not have a solution.
Now if you mean lnx=x^3-1, then yes it has two solutions, Since x3 contains the variable to solve for, move it to the left side of the equation by subtracting x3
from both sides.
ln(x)−x3=−1
The roots of this equation could not be found algebraically, so the roots were determined numerically.
x=0.39044,1
numerically on a graphing calculator?
that would be a choice, I would do it on wolfram.
or
Newton's method:
let y = lnx - x^3 + 1
dy/dx = 1/x - 3x^2
newx = x - (lnx - x^3 + 1)/(1/x - 3x^2)
= (1 - 3x^3 - lnx + x^3 - 1)/(1/x - 3x^2)
= (-2x^3 - lnx)/(1/x - 3x^2)
guess: x = .5
newx1 = .3545...
newx2 = .38789...
nrwx3 = .39042..
newx4 = .39044046
newx5 = .39044046 <---- correct to 8 decimals
looked like some fun to do it.
To solve the equation lnx = x^3 + 1 algebraically, we need to use a logarithmic property known as the inverse property of exponentiation.
1. Start by rewriting the equation as:
ln(x) = x^3 + 1
2. Apply the inverse property of exponentiation:
Convert the natural logarithm on the left side to an exponential equation. In this case, since ln(x) and e^x are inverse functions, we can rewrite the equation as:
x = e^(x^3 + 1)
3. Since we are given x = 1, substitute it into the equation:
1 = e^(1^3 + 1)
4. Simplify:
1 = e^(2)
5. Finally, use the natural logarithm to solve for e:
ln(1) = ln(e^2)
0 = 2ln(e)
0 = 2
Since the equation leads to an inconsistency (0 = 2), there is no real value for x that satisfies the equation lnx = x^3 + 1. Therefore, x = 1 is not a valid solution.